text:
Applying back in the US as foreign with FAA Licenses
Just for context I’m currently sitting at about 250~ hours and applying to airlines back in South America. I did my training in the USA, therefore I have my instrument rating, MEL rating, and Commercial pilot certificate, from the FAA. I converted them back to be able to apply, so all good there.
Is there any chance I can come back to the US with more hours, as all aircraft are “N” registered, and be able to apply for a Job in the USA? If so, what are those options?

FAA certificates do not expire.
You will need a work visa and a company that will sponsor the work visa.
You will need an FAA ATP certificate after you have 1500 hours in registered aircraft. The aircraft do not need to be registered in the USA with N-number, but they cannot be ultralight. If a company is willing to sponsor the visa then they will also organize the ATP certificate tests.
It will be much easier to find a job if you first have an ATP in your home state. This is proof of ability before they spend money to get you an FAA ATP.

Ok, I see yeah I will get my ATP at my home state, I just wasnt sure how were my chances being a foreign and applying back to the US. Thanks for the response

@CES I can only say if you want a job at a company in the USA then send them your application. I do not know your chances, I do know that they do hire some foreign pilots and other foreign employees but I do not know the fraction and it will change between companies and with the economy.
The VISA partly depends on what you want; only working for a few years then returning home or becoming a citizen of the USA.

Thx for the answers appreciate it


text:
CDFA Calculation in Charts
I'm trying to find the accurate formula which is used on a 2D and sometimes 3D approach that calculates the Altitude at which you have to be at a specific distance for a CDFA. Using trigonometry yields a result which is off by 20ft sometimes even more. Is the curvature of the Earth used for this calculations? If you can provide a solution for this example which gives an exact number(if possible) or a closer result from my calculations will be appreciated. I'm starting to believe they round up the values published in the charts.
Ex.
LAX RWY 24R

My calculation to find 2,200' at 6.3nm from rwy threshold:
TDZE= 122'
TCH= 55'
Distance= 6.3nm
Alt= x
x= tan(3) * (6.3 * 6076.12)
x= 2006.1465' + 122' + 55'
ALT at 6.3nm = 2183.1465'... I'm off by roughly 27ft.
Any insight will be appreciated, thank you in advance.
Regards,
mabg

Instrument approaches do include the curvature of the earth in their considerations. At 6.3 nautical miles this is something like 30 feet.
But there are various types of rounding going on, too. Most importantly in this case, the 6.3 nautical miles you're relying on is the sum of 0.9, 0.8, and 4.6, which are all rounded to the nearest tenth of a nautical mile. The FAF is defined as being at 8.3 DME from the localizer, which is not exactly 6.3 from the runway threshold.

It is 6.36 NM from the threshold, which proves your point.

Your maths is correct, though I think you mean you’re 17 feet off, not 27 feet?
Your assumption about rounding is likely correct as well:
Practical Considerations
When considering the chart from a pilot's perspective, especially during a relatively fast approach, a discrepancy of 17 feet at 6.3 NM out is negligible. This minor difference does not impact the accuracy or safety of the approach.
Chart vs Actual Final Approach Path Altitude-Distances
The altitude shown on the chart may differ slightly from the altitude at which the aircraft is actually flying along the final approach path. For example, while the aircraft may be closer to 2183 feet at 6.3 NM, when the pilot cross-checks this against the chart, seeing 2200 feet is close enough. Given that the aircraft is likely descending at around 600-900 feet per minute ($=$ 15 feet per second), this minor difference is not operationally significant.
Rounding Benefits
Pilots are very busy during the approach phase, and having altitudes rounded to the nearest 10 feet simplifies their workload. The aircraft’s instruments typically display altitudes in increments of 10 or 20 feet, so rounding helps in quickly cross-referencing with charted altitudes without requiring unnecessary mental effort.
Altimeter Settings and Altitude Discrepancies
It's also important to note that altimeter settings are typically adjusted to the nearest hPa1, which can introduce a difference of about 27 feet per hPa in indicated altitude. This is another reason why minor discrepancies such as 17ft does not matter in practice. Furthermore, temperature variations can slightly affect the true altitude at a given distance on a final approach path. These small variations are accounted for in the design of charts, which is why rounding is both practical and operationally sound.

To properly answer your question, I believe your mathematical calculations are correct i.e. you are probably being more accurate and precise than the chart. The Earth's curvature will be negligible over 6.3nm. Additionally, if the approach involves an Instrument Landing System (ILS), the glide slope is defined by radio waves, which travel in a straight line relative to the curvature of the Earth. This means that the curvature does not affect the glide slope calculations. The minor discrepancies you observe are more likely due to rounding or simplifications in the charting process rather than any significant geometric considerations.

1.  Some countries use inches of mercury rather an hPa increments, such as: USA, Canada, Mexico and Japan. 0.01 inHg = 10 feet of altitude change

Don't forget that there is are slight variables due to glide slope antenna placement on the aircraft and height of landing gear.

source: Stack Exchange

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sort: votes


text:
Basic question about changing track at flight level
I'm a beginner private pilot flying VFR in Europe. I think I've heard that in the US it's not so common to fly at flight levels on small aircraft, but in Europe the transition altitude can be quite low enough for small planes, and there are definitely places you can fly VFR at FLs.
I want to go out and practice this sometimes (the rule is VFR flights fly at FL+500ft). But there is a basic question I haven't clearly discussed anywhere, which is if I change track such that I have to ascend or descend to a different FL according to the even/odd rule, what is the procedure? Change track first and then descend (for example)? Or descend first? Or do both more-or-less simultaneously?  Though at 500ft/m there will still be about 2 minutes of transition either way. How can this be ensured safely?

Airline pilot here who did all his training in the UK. I could be wrong, but there may be no need to fly at a FL even if above transition alt. The only time I flew flight levels during training was along airways under IFR. VFR would allow you to fly at any altitude want to -  unless you are under radar control, then the controller will instruct you to fly at a particular altitude/FL.

The ICAO rule for VFR cruising altitude is as follows:

Except where otherwise indicated in air traffic control
clearances or specified by the appropriate ATS authority, VFR
flights in level cruising flight when operated above 900 m
(3 000 ft) from the ground or water, or a higher datum as
specified by the appropriate ATS authority, shall be conducted
at a flight level appropriate to the track as specified in the
Tables of cruising levels in Appendix 3.

These cruising levels are altitudes below the transition altitude and flight levels above. There is no difference in the rules except what altimeter setting you should be using.
Unless preempted by local laws, there are no rules about how cruising level changes should be made. You should avoid flying in the wrong direction for the altitude you are flying for an extended period of time.
From a safety standpoint, the primary means of staying safe is "see and avoid." Radar services/ADS-B/TCAS and other technology can help enhance your safety.

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text:
Are all jetliners equipped with cabin air filtration system?
I want to know if all commercial jetliners are now equipped with cabin air filters or not. I knew that, till some recent period, only the cockpit had this equipment and not all aircrafts.

Not all planes have HEPA filters, but since the COVID-19 pandemic, all new airliners and most major airline fleets have been equipped with HEPA filters throughout the full cabin.
So anything that rolls off Boeing and Airbus production lines has HEPA filters for the full cabin since COVID.
Some of the older planes like American Airlines Regional jets don't have HEPA filters but many of these old planes are being retrofitted.
Lufthansa, ANA, Southwest, and many other airlines have all their planes with HEPA filters.

Since 1988 on Airbus single aisle aircraft and since 1994 for the entire fleet (see figure 5), Airbus installs HEPA (High Efficiency Particulate Air) filters for the recirculated air providing more than 99.9% filtration efficiency (for 0.3 micrometer particles as most penetrating particle size). It is important to understand that the filtration efficiency for these filters is not only based on direct interception or sieving for larger particles but also on other effects like impaction and diffusional interception for smaller particles like bacteria and viruses.

I got this information from here
Here's another document from IATA on this matter - Cabin air quality


text:
What is the radius that the METAR and TAF covers around an aerodrome?
Is there a limited radius for a METAR/TAF?


text:
How to know what to insert in the TO SHIFT in airbus MCDU?
I understand that TO SHIFT is used when departing from an intersection to update the position of the aircraft in the lack of gnss. But how do I know what to put in the TO SHIFT? Is it availabe in the ADC?
Thanks in advance!!

You need training on how to use a manual transmission and clutch to use that feature.

The TO SHIFT is entered either in meters or feet depending on your operators selection of units ([M] or [FT] will displayed next to the field). This image from the FCOM shows an example with 900 m takeoff shift:

(Airbus A320 FCOM - Auto Flight - Flight Management - Takeoff Update)
To figure out the takeoff shift for a given intersection, look at the Takeoff Run Available section on the ground chart. Here is an example for London Stansted (EGSS):

If you want to takeoff from runway 04 at intersection K1 you calculate the difference from rwy head:
$$ \text{TO SHIFT} = 3049 \, \text{m} - 2823 \, \text{m} = 226 \, \text{m} $$

thank you very much! It seems not all aerodromes have this info tho. For example, here in Brazil, at SBGO, we don't have the Takeoff Run Available section on the ground chart (at least, I couldn't find it). So, in situations like this, how do we determine what we have to enter in the TO SHIFT?
I understand it very well seeing that some aerodromes have this info available in the ground charts because you have the info of the TORA of each intersection. But I'm confused on what should I do if I don't have this info availabe.

@Cessna152Driver Sometimes the takeoff run available is on a separate chart, but for SBGO my Jeppesen charts indeed do not include any such section. If these are not officially published and you don't get any official values from your operator, I would reject an intersection takeoff clearance. Remember that intersection takeoffs are always optional.

Some airports might not mention it in the charts, but they will have it displayed on a board near the intersection point of the runway. of course, you can always ask the ATC. And if the ATC wants you to go for an intersection they will explicitly mention the TORA.
In aircraft like an A320, you will use performance software tailored by your operator which will have intersections already in it. If you are using the old charts, they will also be tailored for that. I have done it both ways.

Takeoff shift is the Take off Run Available (TORA) minus take off distance available from the intersection.
For example, if a runway has a TORA of 3400 meters and your intersection take off distance is 2600 meters, takeoff shift will be 800 meters.


text:
Why isn't horsepower used as a measure of power output for jet engines?
I don't understand why we can't use hp to measure the power output of a jet engine. Is it because jet engines operate on a different principle from piston engines? What determines the appropriate unit?

Of course you can express the power of a jet engine in horsepower. It is rather impractical, though.
The first step to practical units is to switch to SI units. Power is better measured in Watts, which is equivalent to Newton times meters per second.
Now let's use a fairly common turbofan engine, the JT-15. Its version D has a static thrust at sea level of 12.7 kN. Static means it is motionless, so the power is zero. Better to pick up some speed: At 245 m/s the thrust drops to 8.0 kN. Times the speed this becomes 1.96 MW or 2626 HP.
The purpose of a jet engine is to overcome drag and accelerate an object like an airplane. With 12.7 kN static thrust, you can directly apply Newton's first law when you know the mass of the airplane and determine the acceleration. Say the airplane has a mass of 4000 kg: A JT-15D without significant installation losses will give that airplane an initial take-off acceleration of 3.15 m/s². As you can see, knowing the thrust is more helpful than knowing the power.
Turbojets have approximately constant thrust over the subsonic speed range, so a single figure is all you need to get an idea how well they will power your airplane, regardless of speed. As you can see from the above example, at the same time power will vary wildly over speed, making quick back-of-the-envelope calculations a bit harder.
Why, then, would anyone express engine performance as power? This has to do with piston engines, which are rather unaffected by flight speed and deliver constant power over speed. With them it is easier to express performance in Kilowatts or horsepower, because that figure is now constant over speed. To arrive at the thrust you still need to know the efficiency of the propeller attached to this engine, which in turn is quite variable  over speed.

power depends on the frame of ref. There is also the pumping power (volume times pressure).

@fraxinus ... which would obviously be better for jets. But the thrust times speed sort of power is also used for propeller driven airplanes, necessitating an exception for the static case. And, of course, I wanted to show how much more practical the use of thrust is.


text:
Why do we Brief Jeppesen Charts from right to left?
During my IFR training, we used jeppesen charts. I've been informed by my flight instructors that we start briefing a Jeppesen chart from top right to top left. For example, we start reading the city name and procedure, then the index number and revision date, and then the icao/easa code and airport name. However, I'm having trouble understanding why it's like this. I asked my instructors and none could give me a convincing answer. Why is it not top left to top right? Is there a document that specifies this? I couldn't find it.

What you need to do is to have somebody show you a Jeppesen binder, from the days before iPads.
Imagine you have a book or binder and you're trying to find the right page. When quickly fanning through the pages, the most visible portion is the part furthest away from the binding. This means the top right on right-hand pages. That's where they print the city name and procedure, which you're searching for. Your briefing flow starts there, so it has to go left.
You'll find this universal: Jeppesen, FAA, Lido, NavBlue, UK AIP, etc.
One could start from the top left on left-hand pages, but due to Western left-hand binding, the top right is natural.
You could also use the bottom right, and in fact FAA charts repeat the information there.

This is for the same reason page numbers (and more generally "folios" including page number, running header, important information, etc) are aligned with the outer margin, and sections start on a odd page.


text:
Why does Skin friction drag coefficient decrease with Reynolds number for a flat plate?
I have a question: why does the Skin friction coefficient decrease as Reynolds number increases?
Please see the photo below:

Let’s assume we increase Reynolds number by increasing velocity. Higher velocity = thinner boundary layer = higher shear stresses. Should this not increase the skin friction coefficient?

Friction drag is drag coefficient times dynamic pressure. Of course will drag be higher with higher speed, but the coefficient will be smaller. Only the product of coefficient and speed squared will go up!

Given:

$\tau_w$ is the wall shear stress
$q_{\infty}$ is dynamic pressure beyond the outer edge of the boundary layer

Then:

local skin friction coefficient, $c_{f_x} = \frac{\tau_w}{q_{\infty}}$

$q_{\infty}$ is in the denominator and increases more rapidly with Re than $\tau_w$. For example, if the increase in Reynolds number is solely because of a velocity increase, $q_{\infty}$ is proportional to the square of the velocity. Remember, the Reynolds number is just a ratio of inertial to viscous forces.

Thanks for the edits, @DeltaLima.

Thanks for the edits, @ROIMaison.

You're welcome. If you're interested, have a look at this MathJax tutorial to see how to typeset mathematical equations and symbols on StackExchange.

thinner boundary = higher shear stresses $\tau_w$

Should this not increase the skin friction coefficient?
Not if v$^2$ is in the denominator of C$_f$.
But why?  An understanding might come from a comparison with the interaction of high velocity jet plumes with surrounding air.  Although the boundary layer is thinner, the velocity difference is such that less energy is transferred to the surrounding air per unit velocity, hence a lower coefficient.
This may be due to viscosity being less of a factor as Re number increases.  Once the plume slows down, a thicker, more turbulent layer forms.

"Viscosity" of a $gas$ is a bit hard to fathom, but the lessening of effect with increasing Reynolds number may have a lot to do with velocity.  As the moving surface pulls one air molecule, it will start to pull the next one, then the next one.  (Almost as if they were connected with rubber bands)  If the surface is near the first molecule for a shorter period of time, less surrounding movement (turbulence) will occur.

Please note that that plot should be used in a different way: if you look carefully, the label of the x-axis is $Re_x$ where the "$_x$" actually means: "the local Reynolds number as measured from the leading edge - the "$x$" - of a flat plate".
That plot describes how the local skin friction evolves when moving at a distance $x$ from the leading edge of a flat plate. In particular, it tells us that:

The boundary layer starts laminar by the trailing edge (leftmost part of the plot);
It grows in thickness with a relevant decrease in shear stress and therefore friction (as you've correctly guessed - even if the other way around)
When the distance from the leading edge is such that the relevant $Re_x$ is around $5\times10^5$, then the boundary layer becomes unstable and transitions to turbulent; this transition is accompanied by a jump in the local skin friction (middle line in the plot);
Moving further downstream, the turbulent boundary layer grows again in thickness with a relevant decrease in shear stress and therefore friction (right part of the plot).

Evolution of the boundary layer on a flat plate with the local Reynolds number (picture source)
To get the total skin friction coefficient for a plate of length $L$ you need to integrate that plot from $x=Re_x=0$ (yep, in this case the Reynolds number can be $0$ no matter what the speed is) to $Re_L$ and therefore that result does not depend only on the velocity.

Let’s assume we increase Reynolds number by increasing velocity. Should this not increase the skin friction coefficient?

It should be clear now that the answer is: depends. If the combination of velocity and length is such that the boundary layer remains fully laminar then you are going to get a lower friction than a case where the $Re_x$ is fully turbulent, even if the speed is higher!


text:
How does adding twist to an elliptical distribution alter induced drag? Effect of twist on induced drag
I have been thinking about this for the past few days, and I still cannot get my head around it. So, I know that for a given span, an elliptical circulation distribution will produce the least induced drag due to the constant downwash. However, what if you add washout to a wing which has an elliptical circulation distribution, will induced drag be decreased further?
Downwash wont be constant and your circulation distribution wont be elliptical anymore, but wont downwash be reduced as essentially you are reducing your local lift coefficient compared to the non-twist case?
Like for example, lets say we have a rectangular wing with a lot of wash-in starting from the root towards the tip, and now lets compare it to a rectangular wing with a lot of wash-out starting form the root, like the image below. Which one would have the least amount of induced drag? like can you tell which kind of lift distribution will have lesser amount of induced drag by simply looking at the distribution? I have drawn the examples below:

Shown above is an example of a wing with a lot of wash-out starting form the 'root' aka centre.
If you could help me understand this, it would be greatly appreciated.

Like I am just confused about how to tell what kind of wing has more or less induced drag. Like in the example of the circulation distribution curves I drew, you can see there is an extreme case with lots of washout for a rectangular wing, and an extreme case with lots of washin. which one of these would produce less induced drag?
And is it correct to assume they both produce more induced drag than elliptical because they are not an elliptical distribution?

Induced drag happens when there is a gradient of circulation over span. An infinite wing, where circulation is constant over span, has no induced drag. The higher the gradient, the more the resulting vector of aerodynamic forces is tilted backwards. The higher the circulation, the higher the lift and, consequently, the backward component of aerodynamic forces.
This means you should have a shallow gradient where circulation is high and can progressively steepen the gradient when circulation decreases in order to minimize the backward component of aerodynamic forces, a.k.a. induced drag. That is what an elliptical circulation distribution over span achieves.
The one undisputed advantage of an unswept elliptical planform without washout is an elliptical lift distribution over the full angle of attack range with fully attached airflow. As soon as separation starts, however, that is no longer true. All other planforms need to be tweaked to produce an elliptical lift distribution and then do it only at one angle of attack. This tweaking is done by applying washout.
Ergo, an elliptical wing does not need washout, but when it is applied, induced drag will go up relative to lift.

Ok gotcha, so pretty much you need to try match the elliptical distribution at all points along the span to reduce induced drag. Is it possible to tell which kind of platform produces more or less induced drag relative to eachother? Like in the photo I drew of the rectangular wing with heaps of washout vs washin, how do I know which of these configurations has less induced drag?

@aerokid95 At what angle of attack? Without knowing this, no answer is possible. If you know incidence and airfoi over spanl, do a lifting line calculation. That should be the easiest way to find out.

Like if we assume linear twist from 0 -> +10 degrees for wash-in and -10 degrees for washout. I always thought washout would lower induced drag more than wash-in, because it reduces tip-loading which in my head I thought would help with minimising induced drag…. (Like how modern aircraft run tapered wings). Also, do you have any resources or links that are showing both how circulation and the gradient of circulation vary for different platforms? I would love to visualise this

@aerokid95 Oh, come on. I should not only do the thinking for you, but now also the research? Please grow up.

Sorry! I didn‘t mean to come across like that. I tried searching about circulation gradient over a wing but could not find anything online (only pressure gradient) so I just thought I would ask. Maybe I’m not using the right keywords or whatever. But thanks anyway

@aerokid95 Take a lifting line code and play around with it. This will be much better than any dry explanation.

Everywhere the lift along the span of the wing changes, a vortex is shed perpendicular to the trailing edge that contributes to the formation of a so-called vortex sheet. Well behind the wing, this vortex sheet rolls up into two big vortices that eventually dissipate due to the viscosity of the air. The following picture (from Wikipedia) nicely depicts some of those vortices which roll up behind the wing and the tailplane:

The vortex-sheet is responsible for the generation of a drag component which is termed "induced drag". It can be demonstrated that the induced drag reaches a minimum when the spanwise lift distribution is elliptical; in any other case the induced drag is bigger. In this case of elliptical lift distribution it exists a simple mathematical relation between lift and drag coefficients which is the well known $C_{d_i}  = C^2_l/\pi AR$.
Tweaking twist (aka washout), taper and airfoil shape along the span is how the spanwise lift distribution is modified in order to get as close to the ideal elliptical lift distribution as possible. Unless you have a wing with an elliptical platform, this tweaking is anyway valid only for a small range of AoA.

what if you add washout to a wing which has an elliptical circulation distribution, will induced drag be decreased further?

No, for the simple reason that if you have a washout distribution that already minimizes the induced drag then any modification to it will simply increase the induced drag - by definition of minimum.

can you tell which kind of lift distribution will have lesser amount of induced drag by simply looking at the distribution?

Yes, the elliptical one.


text:
Do pilots have to produce their pilot license to police when asked?
if an airport police officer asks to see the pilot's
license of a private aircraft pilot, does that pilot have to show it?

Legally (as in that they can be arrested for failure to do so) or by regulation (as in they can lose their license for failure to do so)?

In short, yes, although this is in practice very rare and most likely to be the FAA or an NTSB official first (after something has gone wrong) rather than a police officer.
The pilot's license (or certificate in FAA parlance) is just one of several documents that may be requested, depending upon the pilot and what they are doing.
In the US, this comes under FAA regulations; Title 14, Chapter 1, Subchapter D, Part 61, Subpart A, Section 61.51, paragraph (i) of which (!) reads:

Presentation of required documents.
(1) Persons must present their pilot certificate, medical certificate,
logbook, or any other record required by this part for inspection upon
a reasonable request by—
(i) The Administrator [i.e. of the FAA];
(ii) An authorized representative from the National Transportation
Safety Board; or
(iii) Any Federal, State, or local law enforcement officer.
(2) A student pilot must carry the following items in the aircraft on
all solo cross-country flights as evidence of the required authorized
instructor clearances and endorsements—
(i) Pilot logbook;
(ii) Student pilot certificate; and
(iii) Any other record required by this section.
(3) A sport pilot must carry his or her logbook or other evidence of
required authorized instructor endorsements on all flights.
(4) A recreational pilot must carry his or her logbook with the
required authorized instructor endorsements on all solo flights—
(i) That exceed 50 nautical miles from the airport at which training
was received;
(ii) Within airspace that requires communication with air traffic
control;
(iii) Conducted between sunset and sunrise; or
(iv) In an aircraft for which the pilot does not hold an appropriate
category or class rating.
(5) A flight instructor with a sport pilot rating must carry his or
her logbook or other evidence of required authorized instructor
endorsements on all flights when providing flight training.


text:
Why does a Phugoid occur? How can it be eliminated?
My model aircraft seems to do phugoid oscillations and I'm having trouble landing. It is around 2 kilos, very high mounted wing, so bad landing takes out the underbelly everytime.
I'm wondering what's causing the phugoid, CG placement? The extremely high wing causing pitch oscillations due to inertia? Insufficient tail?
How can I solve this? I cannot change the wing configuration, but I can change CG along the fuselage length / modify tail.
Edit: More detail - design:

When does the oscillation happen? In level flight without control input?

@GdD yes in level flight. with inputs, it's larger

Before you do any design changes I'd check your control setup, make sure you aren't getting flutter or play in your control surfaces.

@GdD yeah I've checked them. They're stiff and stable.

it's an interesting question but very broad, possibly too broad for this site. Also, there's not enough detail, making any answer speculation, which is also discouraged. Pictures, designs and any other details would help.

I can add 2d drawings showing dimensions and placement of components. I'm not at home otherwise I could've snapped pictures. But I hope the drawings are helpful enough

Move the center of gravity forward
What you experience looks almost like a phygoid motion (well, most of it is a phygoid), but involves stalling at the low-speed part of the cycle, so it is not the classical Eigenmode.
Rather, your aircraft's trim point is beyond its stall angle of attack. It will, therefore, stall when not actively steered towards a lower angle of attack and then, when lift on the wing diminishes and its center moves backwards while lift on the tail is still linear, the aircraft will pitch down and pick up speed again.
Maybe it will be already enough to change the setting on your elevator to a few degrees more trailing-edge-down deflection. What certainly will help is to move the center of gravity forward because this will increase static stability and require more trailing-edge-up elevator deflection to maintain the same trim point.
From your sketch I would also guess that your tail volume* is on the low end. Move the tail further back by lengthening the fuselage: This will increase stability and, especially, damping, so the pitch motion after stall becomes less violent.

*  This is the area of the horizontal tail surface, multiplied with it's lever arm (distance between center of gravity and the tail's quarter chord point).

So it all starts with the trim point way beyond stall, huh? So if I move the CG forward (which is the easiest modification for me) It'll increase tail length as well, increasing tail volume. I should see if it's sufficient.

@NRB: Yes, start with the center of gravity. But you will only need a small shift, so an extra lengthening of the tailboom is still advisable. Also, you might want to look at the trim point of the elevator: Give it a little more trailing-edge down deflection and see how that works.

The phugoid is an interchange between kinetic and potential energy. Anything that dissipates energy in the process will add damping - more parasitic drag from a speed brake for instance when talking about the landing. It's a bit of a dilemma, anything that is done to clean up the aerodynamic shape of the aircraft will decrease phugoid damping.
Control problems related to the phugoid are caused by the natural frequency being too high, this can be lowered by increasing horizontal tail size: a larger tail for a lower frequency, and more time to anticipate on the motion.
Helicopters have the phugoid behaviour exactly like fixed wing aircraft have, and the military have stability requirements for phugoid behaviour: from MIL-H-8501A
$$\begin{array}{|c|c|c|}
Period & Visual Flight & Instrument Flight\\ \hline
\text{< 5 sec} & \text{½ amplitude in 2 cycles} & \text{½ amplitude in 1 cycle} \\ \hline
\text{5 - 10 sec} & \text{at least lightly damped} & \text{½ amplitude in 2 cycles} \\ \hline
\text{10 - 20 sec} & \text{not double in 10 sec} & \text{at least lightly damped} \\ \hline
\text{> 20 sec} & \text{no requirements} & \text{not double in 10 sec} \\ \hline
\end{array}$$

Near stall speed, the period is around 5 to 6 seconds. If horizontal tail can affect frequency, what will affect the damping (except parasitic drag)? Because the touchdowns are very hard (lots of residual amplitude).

The damping is only affected by $C_L/C_D$. A period of 5-6 seconds is too short for comfort - install a draggy large horizontal tail could be an option.

There can be two factors causing the landing mishaps mentioned:

Phugoid: the natural phenomenon of any aircraft.
Pilot Induced Oscillations: the pilot-in-the-loop oscillation of the aircraft while in control of a pilot.

To test the phugoid, trim the aircraft to a calm condition and do not touch the controls for several seconds.
For improving phugoid (long period) dynamics, especially if it is already causing troubles, a pitch-rate-gyro can assist in augmenting the stability. A sample commercial product here.
The PIO is a far more important problem, and it's a combination of the pilot's behavior (his/her control dynamics) coupled with the airplane's dynamics. During final approach and landing, pilots tend to be more stressed, and if the phugoid is not a lucky one, the touchdown may happen at not the best point.
The gyro would also help with the PIO problem, if any.

It oscillates without intervention for nearly a minute before it's damped completely. I could add gyro assist, but that'll be if I cannot mechanically solve the issue.

Ok, I assume it’s not a CG problem of controllability. still, you could try a forward CG configuration to make your touchdown easier. The phugoid is triggered by (sudden) disturbances (external or control commands), so a forward CG might improve a bit. Else, the gyro would be the way to go.

license: Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/3.0/


text:
What kind of material should I use for rudder skins?
I’ve done some research on this topic for awhile now, but I’m not exactly sure how to go about it. I’ve heard and read that on some World War Two aircraft had fabric covered fuselages, such as the Hawker Hurricane. The Mosquito aircraft was made of plywood and covered with fabric entirely. My primary concern is how the fabric will hold up and interfere with the yaw compared to metal, and I’m also occupied with costs, repair, and maintenance. This question might seem vague to some, but control surfaces are essential to performing maneuvers and controling pitch, yaw, and roll.
The aircraft speeds expected from the V8 engine are a top cruise speed of 193 knots and an economy speed of 170 knots. The rudder situation is an H-tail configuration, single engine powering a four bladed propeller. The aircraft wing is elliptical, in which the wing extends straight out for six ribs, then slants upwards at an angle of 15 degrees. Landing gear is retractable, taildragger configuration. Flaps are plain with a deflection full deflection of 45.

@Dave if doped canvas is suitable for all sizes and speed ranges, does the question really need to be closed for not specifying the size and speed? Voting to keep open.

@Koyovis it looks like the question has been changed anyway so I will remove my close vote.

I don't quite understand why you're worried about the rudder in particular. Having flown a fabric-skinned aircraft, there's nothing special about the rudder compared to other control surfaces.

Some aircrafts have had fabric tails over the years that held up just fine. The Mooney Mite (M18) comes to mind. It topped out around 140Kts but flew just fine with a fabric covered empennage. The Vought F4U Corsair also had a full fabric empennage and operated well in excess of your quoted speeds so there is no problem using fabric for speeds you speak of. The FAA has a decent resource on fabric covering here thats worth reading through.
The problem is less about interfering with yaw and perhaps interfering with everything else. Fabric is not as widely used anymore now that we have engine power to compensate for heavier airframes as well as the ability to work with metals consistently and accurately. Fabric also requires routine maintenance as well as periodic replacement. Fabric has a habit of ripping and needing repairs as well as simply sagging over time causing it to require a potentially full replacement. Since a great deal of aircraft are metal skinned now the people left to work on fabric coverings are slowly dwindling as well.

all_licenses: Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/3.0/


text:
Why is the speed graph of a survey flight a square wave?
Today, December 11, 2024 around 15:00 LT, there was an accident involving a Piper PA-31-350 Chieftain in Victoria, Texas, at the intersection of Zac Lentz Parkway (Texas 463) and Mockingbird Ln. The pilot was the only occupant of the plane and was injured. Four people in vehicles were injured, one critically. The plane was destroyed.
The flight tracking for flight N818BR on FlightAware shows the speed rising and falling. My question is why would this survey flight traveling back and forth show speed changes in a square wave pattern rather than something more sinusoidal (-ish)?
The second image shows the wind speed, gusts and direction from the National Weather Service at Victoria (KVCT) for the time period.

Welcome to aviation.stackexchange.com

The chart shows ground speed.  They're flying at approximately constant airspeed with an approximately constant wind (magnitude and direction).
The square wave depicts the up-wind and down-wind legs nicely.  You would expect a sinusoidal shape if they were flying in a circle.

I played the graph over time using the controls provided and sure enough. during NW flight the speed is at the lowest (the valleys of the wave) and SW flight has the higher speed (the peaks). I don't know why I didn't do that earlier. Just looking at the still image I expect the lows to be during the turns.

From a generic data analysis point of view I immediately wanted to see the heading on the same plot as the speed (and altitude).  I'd also want the wind data plotted on the same timescale (X axis).  It's a shame when all the information is in front of you, but presented in separate graphs - plotting the raw data yourself can be very instructive

I initially wondered if the survey equipment needs to run at a certain groundspeed, but your observation shows that can't be the case.   Good sleuthing !

Just out of curiosity - how do they know the ground speed at all? Is it something sent by the transponder? Do they have a theoretical value calculated by knowing the exact wind speed and direction at the moment - and adding the vector to the plane true speed/direction?

@Mykola I would imagine it's derived from GPS data, simply calculating based on the distance and time between updates.

@Mykola Definitely from GNSS, most likely GPS but possibly also other satellite constellations. For more advanced aircraft the FMS can use even more types of sources. They will be put together with the previous time-steps using Kalman filtering or a similar technique, better not just by differentiating the raw values from each source.

@VladimirFГероямслава, Cadence - Under GPS I meant from any constellation, not just GPS. My question was more about how the data is received by the FlightAware? I thought the transponder is quite simple on such planes, and only sends basic information, not the ground speed which is irrelevant in many cases, and to send the data it must be connected to a sat-nav device, an airspeed sensor would not do. Plus, the Piper in question is old, so, until retrofitted, does not have a sat nav integrated with transponder. I should probably ask a separate question.

@Mykola I actually believe the data is collected by ground based flight following radar.  As you mention, a transponder just squawks an ID (Mode A) and barometric altitude (Mode C).  The position and ground speed is measured by radar.  Radar is good at measuring a bearing (direction from the radar station) and radius (distance from the radar station), but not altitude -- which is why the transponder sends out altitude.  All of this technology predates GPS and other SATNAV technologies.

@Mykola These sites mostly receive data from ADSB. These are not that basic anymore and are connected to the source of position. Also, Transponders with mode S are becoming much more common than those simple mode C and are often obligatory in many airspaces. No need for a radar, an ADSB receiver network is sufficient.

@RobMcDonald I don't think those are collected by radars - I am pretty sure it is ADSB most of the time - I think, most of it collected by the network of volunteers who install a receiver connected to the Internet in exchange for access to the service or just for fun. The radars might be used by airports - but not the web services that the question quotes. Anyway, I asked a separate question and had an answer already, but thank you all for answering.

authors: Vladimir F Героям слава


text:
What is the speed of 1.32 Vmd?

*Aeroplane CL/CD polar curve
 (Image Source: Principles of flight. ATPL ground training series - Author: CAE Oxford Aviation Academy (UK) Limited 2014)
I’m studying Flight Mechanics in my textbook and I can’t figure out the speed of 1.32 Vmd.
I know that Vmd is the minimum drag speed. So, 1.32 Vmd is Vmd * 1.32, but why, according to the screenshot, does this speed have a lower CL and CD?
And where can such a speed be applied?
If someone can help me with this, I would be grateful to you.

Thank you!

I'm guessing it's a rule of thumb for maximum range speed for jet aircraft but have not been able to find the derivation. It's interesting that 1.32 is the reciprocal of the ratio of Vmp (minimum power required for propeller aircraft) over Vmd.

With regards to lower CD, remember that does not imply lower D, since qbar will be about 74% higher at 1.32Vmd than at Vmd.

I suppose you are analysing an airplane with a jet engine. If we consider the usual approximation for the drag coefficient of the entire airplane as being:
$C_d=C_{d_0}+kC²_l$
and we equate the power requested to win that drag with the power available from the engine(s), then we get that:

the speed for best loiter (aka endurance) corresponds to minimum drag (aka best L/D) and it is given by $V_{loiter}=V_{MD}=\sqrt{\frac{2W}{\rho S}\sqrt{\frac{k}{C_{d_0}}}}$

the speed for best range corresponds to 0.866 best L/D and it is given by $V_{range}=\sqrt{\frac{2W}{\rho S}\sqrt{\frac{3k}{C_{d_0}}}}$ which is $\sqrt{\sqrt{3}}=1.32$ times bigger than the previous $V_{MD}$; 1.32 is just the value given in your plot.

Another way of looking at the chart is to put Angle of Attack on the Coefficient of Lift axis.
Higher up represents higher AoA and lower airspeed, hence Vmp and Cl max "above" L/D max and 1.32 V md "below".
Using the lift equation Lift = Area × Density × coefficient of lift × Velocity$^2$, we see higher velocity means lower AoA for the same lift.
AoA is (linearly) part of the coefficient of lift, with the other part being the shape (lifting characteristics) of the airfoil itself.
1.32 V md would therefor be a cruise airspeed for maximum range.


text:
Containerization has revolutionized other modes, why not aviation?
It is obvious the "container" concept (that has revolutionized other modes) as a separate drive train (wings/engines/cockpit etc) and sausage tube filled with passengers is technically feasible. Considering the considerable time wasted embarking/disembarking this has to be an economically sound transition. Why hasn't this concept been applied?

https://en.wikipedia.org/wiki/Unit_load_device

That idea even has been patented: See here. More answers to this question can be found here.

The thing you're missing is people load themselves, so they load in parallel with one flight attendant supervising 10s of people simultaneously finding their seat. Cardboard boxes and baggage require at least one human to actively pick up and load.

Well, and it's even more than that, the human loading process is happening in parallel with many other processes (cargo & food loading, fueling, inspections, crew prep, etc.) so even though boarding is relatively slow, it's not really the limiting factor to how fast the plane can turn around.

Containerization has been adopted in aviation too and it has been done in the same way as in any other segment of transport industry: for cargo, in the form of Unit Load Decives (ULD).
Your question seems to imply that containerization was common when it comes to passengers. It is not. There are no applications in any mode of transportation (land, sea, air) where people would first embark in a cabin which was then attached to a separate system providing the propulsion.
This would, for aviation at least, add considerable amounts of complexity and possible points of failure. Considering the typical aircraft layout,  the cockpit, wings and tail would need a loadbearing "exoskeleton" adding huge amounts of weight. Taking the cabin to the airframe and attaching it to it would need specialized equipment that would need to be available at both ends of the journey.
Also, I cannot see how this would actually save time. People would still have to embark the cabin, and the cargo would need to be onloaded too. The cabin would then have to be transported to the airframe, and be attached to it.

An historical precedent would be the Train ferry, where the passenger carriages are loaded onto a ship without the passengers having to disembark.  However, I agree that this isn't a reasonable model for aviation, where the "train" would just carry passengers from the terminal to the apron.

Train ferry for sure is a close match, I may need to rephrase my answer. However weight considerations would require a specifically made human containers, not quite the same as train ferry.

The passenger part doesn’t have to be structural. Consider the seats on pallets on the QC variants of the 727/737. Just put people on them first.

I don't have the rep to do it myself and not much else to change, so somebody please change Decives to Devices.

"imply that containerization was common when it comes to passengers. It is not."   The passenger railway carriage dating way back is containerization although not intermodal - the transition to HiRail is intermodal although that seems to be more focused on heavy loads to reduce carbon footprint.  Innovation requires creativity not a "can't do attitude".  My question is not "can it be done?" - that is a given - its "why isn't it the norm?

@AnthonyLampard good point on railways you are absolutely correct on that. However, one must remember that innovation is something new and useful. You can bet your behind aviation has the most brilliant minds making design decisions such as this. Containerization simply is not feasible in aviation. Your question "why isn't it the norm" carries the presupposition it was feasible.

@user71659 to be brutally honest seats on pallets seem like a very, very bad idea. Transferring and securing the pallets, extra weight of pallet structure etc. etc...

@Jpe61 The thing is seats on pallets are much cheaper than buying, operating and maintaining 2 completely separate planes, hence the 727 and 737QC, available up to the NGs. It's also done on military aircraft, like the C-17 and C-130, from ordinary seats to full VIP suites.

@user71659 you can't really compare military and civillian solutions  Or well, you can, but it makes no sense. Btw: the F-35 is a containerized human air transport platform. The specialty of this platform is, that the pallet size is one human, and the pallet can be ejected from the acft mid flight, expediting the load delivery.

Containerizing humans has never been done, attempting, or more than fancifully proposed.
Intermodal ISO containers were designed for weight-insensitive modes of transport, such as ships, trucks and trains. They're very durable, built to stand up to the elements, break-ins, and stack ten-high. That also makes them heavy.
You'll notice that most container carriers carry them on the outside: trucks, trains, and partially ships; they're also stored outdoors, exposed to the elements. They have to be tough for that. Airplanes carry cargo on the inside, due to aerodynamics, and don't need it encased in such protection. They aren't as large as ships either.
The only aircraft intended to primarily carry ISO containers was Boeing Pelican, a 3,000-ton GEV/ekranoplan. Even that was mostly for military purposes, as the military favors containers (often as small as 1/3 TEU) for speed and flexibility over fuel economy.


text:
Is the elevator chord included in the calculation of the volume ratio?
I'm a little bit confused, when designing the elevator and rudder, are they included in the calculation of the volume ratio of the tail?
What I mean is after calculating the vertical and horizontal stabilizer size, to determine the control surfaces size I can use the guidelines in the photo, for example if the airplane is GA single and the horizontal stabilizer chord is 20 cm, then Ce=0.45xC=9 cm.
My question is the 9cm of the elevator added to the 20 cm of the tail to have a chord of 29cm, or the elevator chord will be 9cm of the 20cm of the horizontal stabilizer's chord?

Yes, the area of rudders and elevators are included in tail volume coefficient calculations and in your example, the 9cm will be part of 20cm chord. Rudders and elevators contribute to stability as well as control.

That picture is taken from "Aircraft Design: A Conceptual Approach" by "Daniel P. Raymer".
In the text it is clearly stated that those values are in fraction of the whole "tail chord". That means that if from the volume method you get a chord of 1m for your GA airplane then its elevator is going to use the last 45cm of that 1m horizontal tail. You can compare those preliminary values with a real Cessna 172 for example.


text:
Are all TCAS targets on view?
Using TCAS II, are all targets visible? or only when they could get in conflict with own aircraft and suposse a threat?

Seems the same as What is the range of TCAS?, there are subtleties explained in the answer (Diversity of ranges)

Other targets are also available. The details may depend on the aircraft and exact TCAS implementation installed. Here are some examples for the Boeing 777:

TCAS Traffic Display

1 Traffic Targets
Indicates relative position of traffic.

filled red square indicates a resolution advisory (RA)
filled amber circle indicates a traffic advisory (TA)
filled white diamond indicates proximate traffic
unfilled white diamond indicates other traffic

(Boeing 777 FCOMv2 15.10.9 Warning Systems - TCAS Traffic and TCAS Alert Message TRAFFIC Display)
As you can see, also proximate traffic and other traffic may be shown as filled or unfilled diamonds respectively. Proximate traffic is defined as closer than 6 NM and within 1200 ft vertically (or with unknown altitude), but not RA or TA.
Sometimes, it is also possible to filter aircraft shown by relative altitude, e.g.:

TCAS Controls (Transponder Panel)

2 TCAS Airspace Switch
ABOVE – altitude reporting traffic from 2,700 feet below to 9,000 feet above
current altitude displayed.
NORM (normal) – altitude reporting traffic from 2,700 feet below to 2,700 feet
above current altitude displayed.
BELOW – altitude reporting traffic 9,000 feet below to 2,700 feet above current
altitude displayed.

(Boeing 777 FCOMv2 15.10.5 Warning Systems - Traffic Alert and Collision Avoidance System (TCAS))


text:
Are there standards that determine the display of results for built-in tests in avionics?
I have seen GO/NGO on some pages to show the pilot the result of a built-in test. However, in some documents I have also seen TEST OK/TEST FAIL, or perhaps just PASS/FAIL. Is there some ARINC or SAE standard that gives guidance on this?


text:
On average, how long does it take to become competent at performing a go-around?
Some pilots spend one to two hours practicing go-arounds. That seems a bit long. And frequently there will be 2-4 (or more) pilots practicing go-arounds at the same time. The lowest altitude they usually go over the runway is 1000-1400 feet.

Huh?  At most airports the highest you would fly in the pattern is 1000'. Most traffic patterns are set at 800 to 1000 feet AGL. And the purpose is to practice approach and landing, thus actually touching down before going around. Although if you are practicing instrument approaches, you may go around at the missed approach point which is normally about 200' AGL and just short of the runway. BTW, pilots practice takeoffs and landings as that's where small errors can get you in the most trouble.

There's becoming competent, (training) and staying competent, (practice) and both need to happen.  Also, as Gerry mentioned, it is most likely instrument approach training/practice you are noticing from the way you described it.

I personally like to teach students go-around technique on the first flight and have them doing it on every flight thereafter, prior to teaching them landing techniques. Making use of E.L. Thorndike’s Law of Primacy, it’s a good way to ingrain the idea that the student can always go around but landing the airplane is an option.
So to answer your question once or twice on every flight is fine for practice.
Now it’s possible that a pilot may, in addition to go-around techniques, may also be working on workflows and/or memory items during this process, which may require multiple attempts.  I recommend no more than four such attempts, followed by terminating the exercise and landing to digest their perceptions and insights of the experience.
As for minimum altitudes, typically when practicing a low approach, you will not fly lower than 500 ft AGL during the approach and/or to satisfy the requirements of §91.119(a).  ATC at tower controlled airports may restrict you to higher minimums at their discretion.


text:
Are there any plans to make the QF-16 into a drone for actual combat missions, or a fully autonomous fighter?
It appears to a naive layman like myself that the QF-16 is an excellent platform to use for remote controlled missions akin to that of other drones. Except that the F-16 appears to have a broader mission envelope than a Predator or Global Hawk. For one, the F-16 is in general quite capable against the majority of the world's active combat aircraft. As I'm sure anyone qualified to answer knows better than myself, it has a proven record in a variety of air to air missions, dogfighting, and ground attack missions.
Are there any other drones, besides the QF-16 that are even remotely as capable, in comparison, and that already exist, in large numbers, like the QF-16?
The benefits of this should be obvious. The QF-16 is after all a disposable aircraft, yet one that is at least somewhat capable. With a drone, no pilot life is at risk. With an autonomous drone, no pilot is required at all for the actual fight, only planning of the mission. This should be an obvious advantage, and from what little I can tell, various---albeit less capable---drones exploit this advantage already.
Just to provide some more context: Many interesting things can be easily imagined, the F-22/F-35 often talks of using fourth generation aircraft as a "force multiplier" by feeding them data from its sensors, such as target information. Such a force multiplier with autonomous, disposable aircraft seems rather interesting to say the least.
I'd like the know if there are any plans to use the QF-16 as a drone for combat missions, and if not, why not?
I'd also like to know if there are any plans to use the QF-16 as an autonomous fighter, as in, program the target into the aircraft, and send it on its way, and have it return, all without any pilot input.
The last question is the most interesting to me. Is there any publicly known research in this direction? Towards making autonomous combat aircraft? Perhaps the QF-16 platform is greatly underpowered both in terms of hardware and avionics (sensors) to realistically (even if the software could be developed) to act in such a manner.
Are there any plans for such things? If not, why not? Are other aircraft in consideration, or is it in general a bad idea? Why?

There seems to be some ongoing research into just that:

There is an effort underway to develop a UAV version of the F-16. This would be based on the QF-16, the remotely controlled target version of the F-16. The air force is already in the process of converting 210 retired F-16s to QF-16s and it was noted that with a little extra work the QF-16 could be turned into a combat UAV for dangerous missions like SEAD (suppression of enemy defenses) or attacking ground targets guarded by heavy air defenses.

Other research is often related to sixth generation fighters, which are often described as being "optionally manned" meaning that they can fly with or without a crew onboard.
F-16s currently being converted into QF-16s are older models which are likely to require more maintenance per flight hour, and as always it is a matter of capability vs cost vs other platforms. The biggest question for such a conversion is whether or not an imagined MQ-16 would bring capabilities not already covered by existing UAVs to the table, and how expensive it would be to support this.

"The biggest question for such a conversion is whether or not an imagined MQ-16 would bring capabilities not already covered by existing UAVs to the table" Dogfighters capable of pulling >10G turns, since they don't have to worry about killing a squishy human flying around inside them. Also, reduced injuries for your pilots since they aren't getting injured by the extreme G-loads a fighter aircraft can exert during a dogfight.


text:
What is the purpose of these wires close to the wing leading edge of the F-117 nighthawk?
Noticed two pairs of wires that seem to run along/underneath the leading edge of the F-117.
What is the likely purpose of these wires? They exist on the other side as well.

Could be almost anything.  My guess is they are fuel level sensor wires.

Or FBW signals to elevon PCUs.  Nav lights.  Something tells me something to do with the FBW controls or some other kind of very sensitive electronics, because they went to the trouble to separate two small harnesses with all that duplicate Adel clamp hardware when you would expect a single run. So perhaps RF emission interference from each harness was a thing that required them to be separated.

@JohnK they also routed the wires as twisted pairs to minimize rf interference/emissions.

@Jpe61 the other purpose for the separation may just be for damage tolerance from shrapnel, where each run is for one of two FBW PCUs. Same reason you'd have dual control cable runs, one on each side of the fuselage, in an airliner.

It is very likely an analog sensor of some sort (my 50 cents on fuel sensor): cables are too small to bring high currents, and 2 conductors are almost surely an analog signal not a digital one (while 2-wire protocols exist, they would require power and GND wires, so minimum 4 wires for digital).

@mins Yes, solutions exist like PLC and POTS (I2C still require separate power) but are very uncommon in embedded systems design.


text:
mig 21 variable geometry inlet cone specific operating range
what is the specific range of extraction and retraction for the variable
inlet cone of the MiG-21bis?


text:
Why would the tip of a propeller stall if geometric pitch was the same throughout the length of the propeller?
I can't seem to understand this sentence from the FAA PHAK about why propellers are "twisted."
"If the blades had the same geometric pitch throughout their lengths, at cruise speed the portions near the hub could have negative angles of attack while the propeller tips would be stalled"
From what I understand, propeller blades are twisted to have a higher blade angle/pitch angle/AOA at the hub because the central part moves more slowly than the ends. Slower means less "airspeed" means it needs higher AOA/pitch angle to generate the same lift as the tip of the propeller. That makes sense to me.
On the image below, higher airspeeds (like cruise), would decrease AOA of a propeller since the relative wind would be closer to the chord line. If the airspeed is fast enough I see how there could be negative lift at the hub.
At the tips of the propeller though, why would there be a stall? Stalling means the "wing"/propeller reached it's stalling AOA, so that means AOA at the tip must be large. The only way to really increase AOA of a propeller is to increase airflow due to propeller spinning relative to airflow of plane (airspeed). Even so, if the plane is moving forward, why would the tips of the propeller be past critical AOA and stall? Are they saying that at cruise, the propeller is spinning so fast at the tip that despite the increased airflow from airspeed, the AOA increases so much that it would stall? How does this happen unless the plane was moving backwards?

I'd appreciate any input!

Have you taken into account that the "rotational velocity" linearly increases from root to tip? It is 0 at the hub and $\omega R$ at the tip. Btw, I'd have called it "velocity due to rotation"

Yes, that was the basis of the question. See 3rd paragraph where I state this, despite in a more colloquial manner. See selected answer, which addresses the point of the question. Thanks.

The small magnitude of velocity at the root does not dictate whether the section there is stalled or not.
The local angle of attack determines whether the flow is stalled.  The magnitude of velocity at the root helps determine the local angle of attack.
The FAA document is trying to represent a standard diagram from the theory of the blade element method -- but they've added so much color fade and un-related stuff that it actually makes it more difficult to understand in my opinion.
Here is another example of the same diagram -- search for 'blade element theory' to find more.

Unfortunately, this image is rotated and flipped relative to the FAA one, so I've manipulated it to be in the same orientation here....  This makes the labels read like a mirror, but hopefully you can use this to convince yourself that it is depicting the same thing as the FAA version, then we can work with the one where the labels are readable.

It is helpful to talk about velocity around a propeller in three components.  Instead of x,y,z components, we will use cylindrical components.  Think of a cylinder whose axis is aligned with the propeller axis and the direction of flight.  The circular outside of the cylinder corresponds with the tips of the spinning propellers.  We will call the three components of velocity:
Axial -- along the cylinder axis, parallel to the axis of rotation of the prop, aligned with the direction of flight.
Radial -- along the cylinder radius, from the root of the blade towards the tip of the blade.
Tangential -- tangent to the motion of the blade, this direction follows a curved path, like the grooves on a record.
This diagram is drawn in the Axial-Tangential plane.  The radial direction would be coming out of the paper and is not depicted.
The FAA diagram also ignores the $w_a$ and $w_t$ contributions (in cyan).  We can ignore them for now too.
The forward velocity of the aircraft (airspeed) is $V_\infty$.  This is the primary axial component of the velocity.
The rotational speed of the propeller is $\omega$.  It gets multiplied by the radius $r$ to give the tangential component of velocity $u=\omega r$.
Near the center ($r$ is small), the tangential component is small.  Near the tip, ($r$ is large), the tangential component is large.
Together, $V_\infty$ and $u$ add together through vector addition to form $w$, the velocity each blade section 'sees'.
Although every blade section sees the same $V_\infty$, the variation of $r$ from root to tip means that the magnitude and angle that $w$ makes with the plane of rotation varies from very steep at the root (where $r$ is small) to very shallow at the tip (where $r$ is large).  This angle is labeled $\phi$ (it is not given a label in the FAA diagram).
The geometric blade pitch (measured relative to the plane of rotation) is $\beta$.
If we subtract the local wind angle $\phi$ from the blade pitch $\beta$, we get the local angle of attack $\alpha=\beta-\phi$.
So now we can tell that the local angle of attack of a blade section depends on the flight condition ($V_\infty$ and $\omega$) and also on the blade twist $\beta$.
We know that $\phi$ will be large at the root and small at the tip.
If $\alpha$ gets too big, the blade will stall.
The FAA mentioned an un-twisted blade (like a ceiling fan blade).  In such a blade, the twist will be constant $\beta=Constant$.  Lets assume $10^\circ$.  In that case, considering the variation of $\phi$, we are forced to have $\alpha=10-\phi$.  At the root, $\phi$ is large, so $\alpha$ might go negative.  At the tip, $phi$ is small, so $\alpha$ might be large enough to stall the airfoil.
One idea for designing a propeller might be to seek to achieve constant $\alpha$ (real props are more sophisticated than this, but this is good to the first order).  So if you want to achieve constant $\alpha$ (say $3^\circ$), and we know $\alpha=\beta-\phi$, then we will need $\beta$ to follow $\phi$ in behavior -- large at the root, small at the tip.

Thank you! That is very clearly put. So is what I'm missing this: FAA document is trying to make a point, and I am taking it too literally. I was wondering if there were some forces that I missed or weren't discussed to cause this stall at the propeller tips. In an un-twisted blade, the propeller tip could stall. If the aircraft is not moving backwards, even if r is very large, ϕ would be very small but still non-negative. So unless the twist (β) is so large that it's past the critical angle of attack where the stall happens to begin with, the tip wouldn't stall?

That is correct.  If you work with a single value of twist, $\beta$ is probably quoted at 75% of the blade radius.  That value would be small.  However, at a blade root, $\beta$ will certainly be higher than the stall angle of attack.  I believe the FAA document is trying to make a point -- without twist, propeller performance will suffer either at the root or tip (or both).  By twisting the blade, you can try to get the most out of the entire blade.


text:
Why is the normal operating airspeed so much lower than the cruise speed in this table?
I need to work with the following table. I'm interested in vMO which I assume I can simply obtain from converting MMO to vMO through $M = v/ \sqrt{1.4 * 287 * T}$. (Is that correct?)
However, I noticed that the vNO is more than 100 kts less than vcr in this table. From my current understanding, vcr should be given as a TAS, which would mean that the the only way in which vNO could be lower than vcr would be that vNO is given in IAS. However, in that case the difference should be much more than 100 kts, right?

Regarding your first part: $V = M \cdot a$, $a = \sqrt{\gamma \cdot R \cdot T}$ with $R = 287.05$. That computation is correct.
Considering the table: are you sure these speeds are at the same altitude? Similarly: do you have a clarification of what the author of this means with the symbols?

Unfortunately I don't know for sure at which altitude vNO/MMO are specified, neither do I know what the symbols exactly mean, but I'm assuming that these are used according to the standard definitions since the author doesn't provide any other information. We're supposed to use this table for an assignment, but for now it's causing a lot of confusion.

VMO and MMO are fix values independent of altitude. VMO is given in IAS. You can convert MMO to an airspeed based on altitude, but really they are two independent limits, and whichever limit is lower at your altitude must be observed.
Your max cruise speed data only makes sense if a TAS; converting it to a Mach number at the given cruise altitude reveals it is the same as MMO only expressed in TAS at given altitude.

Is vNO given in KIAS as well?

@Daniel It has the same values as VMO, so yes. It is normally referred to as VMO as far as I‘m aware.

You really answered your own question. Vne is based on IAS, the actual aerodynamic pressure at any given altitude and speed.  This is why the Mach number is included too, as approaching transsonic adds additional stresses to the airframe.
TAS, as mentioned in previous threads will be much higher than IAS at 35000 feet.  It is fun to enter speeds and altitudes into the TAS calculator to see how close one gets to Mach 1 at 250 IAS when you climb.
TAS is groundspeed without wind speed and direction factored in.  It is the actual aircraft airspeed relative to the airmass. Notice 480 is Max Cruise.

The numbers don't really match up though. For instance, if I enter 31000ft, 1013.25 hPa,-60°C and 335 kts in here https://www.dauntless-soft.com/products/Freebies/TrueAirspeedCalculator/, then I get 541 kts, which considerably higher than the 480 kts vcr. But maybe that's just due bad quality of the data?

Also, to make sure, vNO and vMO are also given in IAS?

So, you can cruise at a lower IAS and save some fuel, no?

That's what going high does for you, as long as you stay within subsonic limit.

"So, you can cruise at a lower IAS and save some fuel, no?" So basically you are saying that if one follows the vNO restriction at this altitude, vcr_max is exceeded? So basically at cruise the limiting factor is the max cruise speed instead of vNO?

Even the spaceshuttle flew 250KIAS on its reentry. Whilst doing Mach 28 of course :)

@Daniel Vno restriction is UPPER limit.  I'd go all the way down to Vy and fly higher if they'd let me. Cruise speed listed in TAS so pilot/engineer can do fuel calc after wind factored in.

As you go higher, Mach number and stall IAS converge, but airliners do not come as close as the U-2!


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What's the aircraft painting standard?
Why are wings always painted grey, regardless of the color of the fuselage? Is there a regulation for that issue?

Just a suspicion: Painting a plane costs money and the paint adds weight. The wings aren't nearly as visible* to the paying public as the side of the fuselage or the vertical stab, so there's less benefit from spending the cash and adding the weight, therefore it's not done. *About the only time you see them is when you're in the terminal, standing at the windows and looking down at the plane. On my last several trips, the only people doing that were under the age of 10 and they don't have much purchasing power.

Your assumption isn't correct, grey is a common color but not the only one wings will be painted.

@FreeMan The wings on planes owned by every airline I just looked at just now were definitely painted and not "bare" aluminium (which would actually be clear-coated but that's still less weight than paint plus clear-coat).

@DavidRicherby valid point. I was thinking "painted with company logo/advertising" not "painted to be finished" as is indicated in the duplicate (for with I voted).

Most of the commercial airplanes I see going overhead (I'm about 10 miles from the approach end of the runway) are painted white, not grey, at least on the bottom.  And the top side of the wings is about the only part of the outside of a plane that a passenger can see.


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On a bypass engine does the bypass air travel faster or slower than core air?
Using the convergent /divergent duct theory the low pressure cold stream bypass air would travel faster but I read on a web page that the bypass air travels much slower than core air. But again using the convergent / divergent duct theory core air is subject to higher pressures and temperatures therefore should theoretically travel slower?
Or is it because air is mixed with fuel and accelerated during combustion which makes the core speed faster and Bernoulli converging / diverging duct theory I stated above doesn't apply?
Hopefully someone can tell me if I'm thinking this incorrectly... Simple explanation please and not something over complicated.
Thanks!

The simple way to think of it is the bypass engine is kind of part way between a pure jet and a turboprop.   In a nutshell, the pure jet moves a little bit of air a lot; the turboprop's propeller moves a lot of air a little, relatively.  In other words, small mass at high velocity vs large mass at low velocity, to get the same push.
A bypass jet engine is somewhere in between.  It's just a fixed pitch non-geared turboprop with 30 or 40 blades, in a shroud, instead of 3 or 4 variable pitch blades in the open (geared turbofans get even closer to the "fixed pitch turboprop" thing).
The package of air being accelerated is smaller than a propeller, but much larger than a pure jet, so fan's exhaust velocity is also in between; faster than the wash of a propeller, slower than the exhaust of the core.

30 - 40 might rather be 20 - 30/40. GE90-fan, for example, is 24, if I'm not mistaken.

@Bram GE90 fan has 22 fan blades.

Bernoulli’s principle, as we refer to it today, states that “as the velocity of a fluid increases, the static pressure of that fluid will decrease, provided there is no energy added or energy taken away.”

(source)
As that assumption is invalidated (energy is transferred, precisely because the air is sped up), Bernoulli's principle doesn't apply. That's as simple as it gets.
A little more detail: when the fuel-air mixture is burned inside a gas turbine, it expands and drives one or more turbines. These are connected to one or more compressors, which pressurize the air that is fed to the combustion chamber. (This is why a jet engine cannot start itself.) The shaft that drives the compressors also drives the fan, which not only does some compression on the air that goes to the core, but also accelerates the air that goes past the core, through the bypass. Nevertheless - and per what you've found online - that air travels much slower than the jet exhaust, as it sees much less compression than the core air. The schematic below illustrates this.

This answer (from where I copied the schematic above) about what part of the engine generates most trust also discusses speeds.

Over most of the engine length, the bypass air travels at the same speed but is much faster than the core air entering the combustor
Let's assume a typical fan engine like the CFM56 pictured below. It starts with an intake which helps to equal flow speed regardless of aircraft speed. The first stage is common for both the core and the bypass stream and increases air pressure without accelerating the air. Next, the flow is split into the core and the bypass stream. The core stream is compressed further and again the pressure rises while speed stays roughly constant. Before entering the combustor, the air is slowed down substantially so combustion is actually possible within the engine.

Cutaway view of a CFM56 fan engine (picture source)
In contrast to that, the bypass stream continues at roughly the speed at which it entered the fan. This is illustrated by the constant cross section of the bypass duct in the picture above. Only at its end does the duct narrow in order to convert the air pressure into speed. Doing this any earlier would mean higher friction losses.
In contrast to that, the core air is slowed down in the diffusor and picks up speed again as it is heated in the combustor. With the higher speed of sound of the heated air, it now becomes faster than the bypass air as it travels through the turbine stages. In the nozzle it is accelerated again in order to convert the remaining pressure into speed.
Therefore, the gas stream leaving the engine core is much faster than the bypass stream, but has gone through a deceleration and then an acceleration while the bypass stream never got decelerated.

The convergent/divergent duct theory that you refer to only applies if there is no energy added. But by definition, this is what an engine does.
Adding energy to the airstream can result in higher pressure or velocity, and this is indeed influenced by the shape of the flow channel. But the hot stream has an internal energy that is seriously higher than that of the bypass flow.


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Why does Beijing's new Daxing airport have so few gates?
There is a lot of press about the Daxing airport, touting that it will be one of the largest in the world in terms of passenger capacity. The images of the terminal build are indeed beautiful.
But when I look at this image, taken from Wikipedia, there doesn't seem to be a ton of gates:

I couldn't find an authoritative number of how many gates it will have, but this website says it will have about 80. This matches what I see in the image above.
By comparison, Ohare and Hartsfield–Jackson have about 190 gates, JFK 130 gates, Schiphol Airport 165.
Why is this? In the Beijing Capitol Airport right now, there are not enough gates, so when I land there, they have us deplane on the tarmac and take a bus to the terminal. Why didn't they build 200 gates if this is to be the largest airport, by passenger volume, in the world?

Not sure if this is relevant here, but 10 A380 gates can possibly have more capacity than 20 gates for A320s/B737s... This is mitigated by the fact that larger planes usually have a longer turnaround time, but it shows that passenger capacity and number of gates are not necessarily directly correlated.

The Wikipedia page says 79 contact aircraft stands.

A380s are no longer being produced - bad clash of between that and terminal design? Seems more gates for smaller planes will now become a driver.

It’s worth noting that ‘having enough gates’ and ‘deplane on the tarmac’ are not mutually exclusive.

@Jan For a couple of years now, every time I land at PEK we can't walk to the terminal, instead we must take a bus. This is after about 6 or 10 expensive overseas flights. I asked the staff "why?" they said "not enough gates". This was for United, Hainan, China Southern, and Air China. If you know something I don't, please expand a bit.
I honestly can't imaging flying from OHare to Amesterdam and not walking off the plane directly to the terminal for a couple years

@Jan There was a Delta flight direct from Detroit that also deplaned without a gate. After 13-15 hours on a plane, traveling with kids, it is a pain to spend an extra 30 minutes getting to the terminal.

Just for fun: At Hong-kong incoming you can get a Pearl River Ferry gate :-) - I've done it once. The ferry is still in international space and customs clearance is at a wharf at Foo Yung - very near Shenzhen's Baoan airport. [You get short HK airport bus ride BEFORE customs to the fery wharf at HK. The first time you know if they successfully intercepted your bags is when they open the large flat bag containers at Foo Yung. If they are not there there is no way to readily find them. Mine were :-).

There are a couple of things I would mention that together could constitute an answer.

79 gates isn’t all that few. For comparison, Osaka-Kansai has 53 gates in its two terminals and Tokyo Narita 97 spread across three terminals (not including bus gates). Tokyo Haneda has only 63.
While currently all international and most domestic travel to Beijing is handled by Beijing Capital airport, in future Daxing and Capital will operate alongside each other. StarAlliance is set to remain at Capital, SkyTeam will move to Daxing, and OneWorld and non-alliance aircraft can choose—some want to serve both airports others only one. Since Daxing isn’t intended to serve all of Beijing, it doesn’t need to have gates for all of Beijing.
While it is expected that a second fully international airport will free up jetbridges at Capital airport, you cannot just expect all flights to be serviced by them. Bus gates exist and will continue to exist. Some airlines—low cost carriers were explicitly mentioned—will choose to use a bus gate rather than a jetbridge for any reason. So while a higher percentage of bus gate arrivals may support the assumption that an airport is overcrowded, it does not necessarily mean that.

"Some airlines—low cost carriers were explicitly mentioned—will choose to use a bus gate rather than a jetbridge for any reason." – RyanAir, for one, would still not like gates even if they were gratis. If they can shave just one minute off their turnaround time by not needing to push back (for example), they will gladly take the outfield two cities over. Being able to roll in, and roll straight out again without the need of a tug is a huge timesaver. (They don't do this everywhere, but they like to do it wherever possible.)

@JörgWMittag I see no contradiction.

@Jan As I stated above, these were not low cost, short flights. These included United and Air China.  Not trying to detract from your answer, I just want to be clear that what I have been experiencing in PEK is different than the Ryan Air low cost thing.

I also want to point out that the Daxing airport is touted to become the largest airport by passenger traffic in the world. That means the intent, according to the press, is that it will be bigger than the PEK. I want to know how this can be accomplished with less than half the gates to ATL. A good answer would have sources as well, not jut speculation or logic.


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Helicopter attitude during take-off and landing (wheels, not skids)
Helicopters with wheels can take-off and land in a level or nose-up attitude (nose wheel off with main wheels on the ground). For example, the pilot of a VIP helicopter might want to take-off in a level attitude for max comfort of the passengers (some nose down will eventually be needed though for acceleration).
Per the above, is there a preferred attitude for take-off and landing of helicopters with wheels or is the attitude only based on performance and design parameters such as CG location, rotor downwash on aft fuselage, rotor mast angle, etc.?

Inverted attitude take-offs are generally frowned upon

Don't forget inverted landing.

Helicopters with wheels tend to be the heavier ones, whose downwash while hovering near the ground is strong enough to be dangerous ("goes out and up hundreds of feet and can easily overturn parked aircraft"), says a blurb offering wheeled tugs for skidded craft.
So when taking off or landing, the longer you stay in translational lift, the briefer you need that dangerous full hover power.  That argues for nose down soon after takeoff, or even where possible a rolling takeoff like a gyrocopter.  Similarly when landing, prefer a rollout ("two-point landing" on the rear wheels), or at least not a long purely vertical descent.

Thanks a lot Camille, just a quick question to clarify; when downwash is an issue, nose up first at take-off and main wheels down first at landing right?

For forward flight (so you can see where you're going), at takeoff nose down as soon as possible to quickly reach translational lift.  At landing, nose up aka main wheels touch first, to stay longer in translational lift.

In a conventional helicopter (main rotor plus tail rotor) the main rotor is normally tilted some 3 to 5° forward. This is done in order to  give the thrust a forward component which partially counteracts the aerodynamic drag encountered during level flight. In this way the main rotor mast is relieved of some structural stress.
The attitude of the fuselage is normally chosen in such a way that in level flight it is horizontal. That means that the main rotor mast  is not perpendicular with the floor but at some 85 to 87°.
When designing the landing gears then two choices arise:

either you set the floor parallel to the ground - and therefore the main rotor mast remains tilted forward 3 to 5°; or
you set the main rotor disk parallel to the ground - and therefore the floor remains tilted backward 3 to 5°.

If you select the option 1. then at takeoff as soon as the main rotor starts generating enough thrust to lift the helicopter, the forward component of the thrust (due to the 3 to 5° forward tilting) starts to push the helicopter forward: if you have skids, this forward sliding would obviously generate a lot of scary sparks and uncomfortable screech... That's why any helicopter with skids is built according to the solution 2.: on ground the mast (and the thrust) is perpendicular to the ground and the floor is tilted backward a couple of degrees. The H145 is a good example of that (picture source, plus a couple of lines from my side):

If your helicopter has landing gear with wheels, then solution 1. is convenient: being the main rotor tilted forward, as soon as thrust is generated the forward component pushes the helicopter forward giving it the possibility to move along the landing field.

That's not what I was after sophit but thanks for the detailed reply that had very useful info!

The second part of your question is more correct.  The pilot doesn’t “choose” a fuselage attitude by rocking the aircraft forward or back under the rotor.  There isn’t a control that does that.
As the pilot picks up, the weight of the aircraft transitions from the landing gear (whether skids or wheels) to the rotor disk.  So sitting on the ground, the attitude of the fuselage is whatever the lander gear configuration dictates.  And once picked up, the fuselage hangs under the rotor mast at whatever angle its fairly narrow CG range dictates. The pilot doesn’t “select” or control the fuselage attitude.
The control inputs the pilot has is to move the cyclic forward…  which increases the lift in the back of the rotor disk, which necessarily moves the helicopter forward over the ground.  If continued this will gradually tip the fuselage forward, as it is a pendulum hanging under the disk that is now accelerating forward.
If the pilot pulls back on the cyclic, it increases the lift at the front of the rotor disk, which tips the rotor disk backward, which begins movement to the rear.  Again if continued, the fuselage will gradually start to tip back, as a pendulum would.
But the fuselage angle is just a natural torsional and aerodynamic consequence of the changing rotor angle.  The fuselage will naturally lean or tip into the direction of movement, and the pilot is controlling the movement, not the fuselage deck angle.